1. **Problem Statement:** Find the area of trapezoid PQST with bases $TS=12$ and $PR=24$, and legs $PT=10$ and $SR=10$. 2. **Formula for area of trapezoid:** $$\text{Area} = \frac{(\text{base}_1 + \text{base}_2)}{2} \times \text{height}$$ 3. **Step 1: Identify bases and legs:** - Top base $TS = 12$ - Bottom base $PR = 24$ - Legs $PT = SR = 10$ 4. **Step 2: Find the height:** Since legs are equal, trapezoid is isosceles. Drop perpendiculars from $T$ and $S$ to $PR$ at points $X$ and $Y$ respectively. Let height be $h$. 5. **Step 3: Calculate the height using Pythagoras:** The difference of bases is $24 - 12 = 12$. Each leg forms a right triangle with half the difference of bases as one leg: $$\text{half difference} = \frac{12}{2} = 6$$ Using Pythagoras theorem on triangle $PTX$: $$h = \sqrt{PT^2 - 6^2} = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8$$ 6. **Step 4: Calculate area:** $$\text{Area} = \frac{12 + 24}{2} \times 8 = \frac{36}{2} \times 8 = 18 \times 8 = 144$$ **Final answer:** $$\boxed{144}$$